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I would think the benzene anion (A) is more stable b/c the compound aromatic. Yes I understand that sp hybridized carbons are more stable than sp2 for negative charges, however I thought that aromaticity is a more important factor in determining stability.
Aromaticity enters the reckoning only if the excess electron associated with the negative charge enters the conjugated pi electron system.
In your case that does not happen. The negative charge is on a formally nonbinding pair on the carbon like other carbanions, but in a conjugated ring when you do not have a ligand on that carbon the electron pair takes the place of the ligand and isn't in the pi system. So not stabilized by aromaticity. Here you don't have the ligand, the top carbon in A where they put the negative charge has no hydrogen atom. If a hydrogen atom were there the molecule would be neutral benzene.
You can have a negative charge stabilized by aromaticity in two ways represented by the benzyl anion and the cyclopentadienyl anion.
Benzyl anion: This has the formula $\ce{C6H5CH2^-}$, a phenyl ring attached to a methylene group. The methylene group acts as the necessary ligand that's missing in your $\ce{C6H5^-}$, so the excess electron pair enters the pi system and conjugates with the ring. Note that the $4n+2$ rule for aromaticity does not work because with that extra pi-bonded carbon you don't have all the conjugated electrons in a ring anymore. But you do have a highly de-localized structure that inherits the stability of the ring.
Cyclopentadienyl anion: this ion, $\ce{C5H5^-}$, has the negative charge and all the ligands on the ring atoms, so the extra electron pair and negative charge are already in the pi system without having to conjugate with anything else like the case above. You have a pure aromatic ring with the negative charge built in. Pyrrole, furan, and thiophene are a variation on this theme where the ring is negative but neutralized by a positive formal charge on the nitrogen, oxygen or sulfur atom.
